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Young's inequality states that if $a, b \geq 0$, $p, q > 0$, and $\frac{1}{p} + \frac{1}{q} = 1$, then $$ab\leq \frac{a^p}{p} + \frac{b^q}{q}$$ (with equality only when $a^p = b^q$). Back when I was in my first course in real analysis, I was assigned this as homework, but I couldn't figure it out. I kept trying to manipulate the expressions algebraically, and I couldn't get anywhere. But every proof that I've seen since uses calculus in some way to prove this. For example, a common proof is based on this proof without words and integration. The proof on Wikipedia uses the fact that $\log$ is concave, which I believe requires the analytic definition of the logarithm to prove (correct me if I'm wrong).

Can this be proven using just algebraic manipulations? I know that that is a somewhat vague question, because "algebraic" is not well-defined, but I'm not sure how to make it more rigorous. But for example, the proof when $p = q = 2$ is something I would consider to be "purely algebraic":

$$0 \leq (a - b)^2 = a^2 + b^2 - 2ab,$$ so $$ab \leq \frac{a^2}{2} + \frac{b^2}{2}.$$

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By the way, I wasn't quite sure how to properly tag this question. Apparently proof is not allowed. –  asmeurer Dec 16 '12 at 7:22
    
I think fundamentally, there is no algebraic definition of $x^p$ in general when $p,q$ are real. But Ben's proof is good for when $p,q$ are rational. –  Thomas Andrews Dec 16 '12 at 8:14
    
@ThomasAndrews yes, I know that, but it still satisfies certain algebraic properties, which I would take for granted. For example, conditions on $x$ or $p$ for $x^p < x$ to hold. –  asmeurer Dec 16 '12 at 8:25
    
@asmeurer I've retagged. The original tags you chose are a bit too specialized. The closest I can find to a tag that match your interpretation of "algebraic" is (algebra-precalculus), where the tag-wiki explicitly mentions symbolic manipulations, which is probably similar to what you had in mind. –  Willie Wong Dec 17 '12 at 16:41

4 Answers 4

up vote 18 down vote accepted

This proof is from "Mathematical Toolchest" published by the Australian Mathematics Trust (image).

Example. If $p$ and $q$ are positive rationals such that $\frac1p + \frac1q = 1$, then for positive $x$ and $y$ $$\frac{x^p}p + \frac{y^q}q \ge xy.$$

Since $\frac1p + \frac1q = 1$, we can write $p = \frac{m+n}m$, $q = \frac{m+n}n$ where $m$ and $n$ are positive integers. Write $x = a^{1/p}$, $y = b^{1/q}$. Then $$\frac{x^p}p + \frac{y^q}q = \frac a{\frac{m+n}m} + \frac b{\frac{m+n}n} = \frac{ma + nb}{m + n}.$$

However, by the AM–GM inequality, $$\frac{ma + nb}{m + n} \ge (a^m \cdot b^n)^{\frac1{m+n}} = a^{\frac1p} b^{\frac1q} = xy,$$ and thus $$\frac{x^p}p + \frac{y^q}q \ge xy.$$

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That assumes $p,q$ are rational, of course. –  Thomas Andrews Dec 16 '12 at 8:08
    
Cool. I didn't know of this inequality at the time, so it wouldn't have occurred to me. Apparently Young's Inequality actually is a special case of the weighted AM-GM inequality. –  asmeurer Dec 16 '12 at 8:19
    
Which also means that one just has to believe the weighted AM-GM inequality for arbitrary real weights. Or one could use a limit/density argument to extend the inequality for rationals to arbitrary reals. –  asmeurer Dec 16 '12 at 8:23
    
This shows that Young's inequality is equivalent to the weighted AM-GM inequality. It doesn't show how that is proven for rational weights from $\sqrt{ab}\le\frac{a+b}{2}$. –  robjohn Mar 11 '13 at 6:07
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@robjohn What do you mean? All that's used is plain old AM–GM? –  Ben Jan 27 at 5:25

Yes, at least for rational $p, q$. There is a general statement here, which can be summarized as follows:

Every polynomial inequality is a consequence of the trivial inequality $x^2 \ge 0$.

In more detail, to prove Young's inequality for general $p, q$, it suffices by a continuity argument to prove it for $p, q$ rational. By making an appropriate substitution of the form $a = x^n, b = y^m$ where $n, m$ are even integers, Young's inequality for a fixed choice of rational $p, q$ becomes equivalent to the statement that a certain polynomial with real coefficients always takes on non-negative real values when fed real inputs.

By Artin's solution to Hilbert's 17th problem, a polynomial with this property is a sum of squares of rational functions. An expression of this polynomial as a sum of squares of rational functions constitutes a proof of the corresponding case of Young's inequality from repeated application of the trivial inequality.

I don't see how you could avoid analysis for irrational $p, q$, since you can't even define the relevant functions without analysis.

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I don't understand your statement "Every polynomial inequality is a consequence of the trivial inequality $x^2 \geq 0$." This seems to contradict the following statement in the wiki article you have cited. "The formulation of the question takes into account that there are polynomials, for example $$x^6 + x^4 y^2 + x^2 y^4 - 3x^2 y^2 z^2$$ which are non-negative over reals and yet which cannot be represented as a sum of squares of other polynomials, as Hilbert had shown in $1888$ but without giving an example: the first explicit example was found by Motzkin in $1966$." –  user17762 Dec 16 '12 at 9:02
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@asmeuer: it is not necessary to find them to prove that they exist, which Artin showed. –  Qiaochu Yuan Dec 17 '12 at 3:42
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I thought the point here was to find them to prove the inequality. Artin's theorem tells us that our search is a reasonable direction to take (and that purely algebraic proofs are possible, at least modulo choosing rational exponents only). –  asmeurer Dec 17 '12 at 20:22
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On this topic, see also andrescaicedo.wordpress.com/2008/11/11/275-positive-polynomials –  Andres Caicedo Dec 23 '12 at 6:29
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@AndresCaicedo excellent blog post! –  asmeurer Dec 24 '12 at 4:46

It might be essentially easier to write $x=a^p,y=b^q$ and $t=\frac 1 p$. Then you want to prove the inequality:

$$x^{t}y^{1-t}\leq tx + (1-t)y$$

for $0<t<1$ and with equality only when $x=y$.

First, we prove the general AM-GM for any $2^k$ variables. The case $k=1$ is the obvious case:

$$(x+y)^2-4xy = (x-y)^2\geq 0$$

With equality only when $x=y$.

Now assume we have proven AM-GM for $n$ terms, we will prove it for $2n$ terms.

If $x_1,...,x_{2n}$ are real, assume they are in linear order. Then:

$$\begin{align}\sqrt[2n]{x_1...x_{2n}} &= \sqrt{\sqrt[n]{x_1...x_n}\sqrt[n]{x_{n+1}...x_{2n}}}\\ &\leq \frac{1}{2}\sqrt[n]{x_1...x_n}+\frac{1}{2}\sqrt[n]{x_{n+1}...x_{2n}} \\ &\leq \frac{1}{2n}(x_1+...+x_n)+\frac{1}{2n}(x_{n+1}+...+x_{2n}) \end{align}$$

Since we linearly ordered them, the equality holds only if all the $x_i$ are equal. (Check yourself here.)

So, by induction, the AM/GM applies to any set of $2^k$ variables. If $t=r/2^k$, we can choose $x_1=x_2=..=x_r=x$ and $x_{r+1}=...=x_s=y$. Then $$\sqrt[2^k]{x^ry^{2^k-r}}\leq \frac{r}{2^k} x + \frac{2^k-r}{2^k} y$$

Which is just $$x^ty^{1-t}\leq tx + (1-t)y$$

(That was just Ben's argument above, but restricted to the $2^k$ case.)

Since the set of $t$ of the form $r/2^k$, $r,k\in\mathbb Z$ is dense in $(0,1)$, you have this inequality everywhere (although density does not obviously let you conclude that equality only occurs when $x=y$ for the arbitrary real $t$.)

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This does prove the inequality for the case of $t:t\,2^k\in\mathbb{Z}$. (+1) –  robjohn Mar 11 '13 at 6:10
    
Yeah, to get for all rationals, you need the full AM/GM. And to get for all real $t$, you need continuity and the density of the rationals in the reals. @robjohn –  Thomas Andrews Mar 11 '13 at 6:13
    
I think my proof works for all rationals. It uses a pretty simply proven extension of Bernoulli's Inequality. –  robjohn Mar 11 '13 at 6:19

With $\dfrac1p+\dfrac1q=1$, $u=x^p$, $v=y^q$, and $1+pt=u/v$, the following are equivalent: $$ \begin{align} xy&\le\frac{x^p}{p}+\frac{y^q}{q}\\ u^{1/p}v^{1/q}&\le\frac{u}{p}+\frac{v}{q}\\ (u/v)^{1/p}&\le\frac{u/v}{p}+\frac{1}{q}\\ (1+pt)^{1/p}&\le\frac{1+pt}{p}+\frac{1}{q}\\ 1+pt&\le(1+t)^p\tag{1} \end{align} $$ Where $(1)$ is the rational version of the Bernoulli inequality, proven below.


Using the integral version of the Bernoulli Inequality, proven at the end of this answer, we get that for $x\gt-n$, $$ \begin{align} \frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^n} &=\left(\frac{(n+x+1)n}{(n+1)(n+x)}\right)^{n+1}\frac{n+x}{n}\\ &=\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1}\frac1{1-\frac{x}{n+x}}\\ &\ge\left(1-\frac{x}{n+x}\right)\frac1{1-\frac{x}{n+x}}\\[8pt] &=1\\[8pt] \left(1+\frac{x}{n+1}\right)^{n+1} &\ge\left(1+\frac{x}{n}\right)^n\tag{2} \end{align} $$ where the inequality is strict if $x\ne0$ and $n\ge1$.

Applying induction with $(2)$, we get that for $x\gt-m$ and integers $n\ge m$, $$ \left(1+\frac{x}{n}\right)^n\ge\left(1+\frac{x}{m}\right)^m\tag{3} $$ Letting $t=\frac{x}{n}\gt-\frac{m}{n}$ and taking $m^\text{th}$ roots gives $$ \left(1+t\right)^{n/m}\ge1+\frac{n}{m}t\tag{4} $$ Note that $(4)$ is trivially true for $-1\le t\le-\frac{m}{n}$. Thus, for all $t\ge-1$ and rational $p\ge1$, $$ \left(1+t\right)^p\ge1+pt\tag{5} $$ where the inequality is strict if $t\ne0$ and $p\gt1$.

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